Solving Problems

In this essay, I will solve two problems from our textbook Elementary and Intermediate Algebra; I will solve problem 56 on page 437 and problem 10 on page 444. For my first problem, I will choose an appropriate variable to help solve the equation, for my second equation I will identify the form of the equation I end up with once it is solved. I will also introduce five math vocabulary words, they are, extraneous, proportion, cross multiply, and extreme-means and the will be in bold.

Problem number 56 states: To estimate the size of the bear population on the Keweenaw Peninsula, Conservationists captured, tagged, and released 50 nears. One year later, a random sample of 100 bears included only 2 tagged bears. What is the conservationist’s estimate of the size of the bear population? This is a ratio equation and my variable will be b, b= bear population b =100

50 2
2*b= 50*100 cross multiply the extreme-means
2b=5000
b= 5000
2
b= 2500 bears
2500 is the conservationist’s estimation of the bear population. This equation is not an extraneous solution because the denominator does not equal zero.

Problem number 10

y-1= -3 this problem is a proportion
x+3 4
(y-1)*4=(x+3)-(-3) cross multiply the x and y which are the extreme and means 4y-4=-3x-9 distribute the 4 and the -3
4y=-3x-9+4 add 4 to both sides
4y=-3x-5 divide both sides by 4
y= -3 x-5 both of these fractions are in parentheses the x and – are outside the () 4 4

The linear equation in the form of y=mx+b with the slope of -3, 4

This essay discussed two problems from our textbook Elementary and Intermediate Algebra. It introduced five vocabulary words to help explain the steps in solving the two problems, those vocab words are extraneous, proportion, cross multiply, and extreme-means. This essay identified the variable in problem one, and identified the form of the equation in problem two.

Reference
Dugopolski, M. (2012). Elementary and intermediate algebra (4th ed.). New York, NY: McGraw-Hill Publishing.

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